TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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A third derivation of $(n + 1) C_{n} = (\binom{2 n}{n})$ , from the closed form of $C (x)$

Statement

For every $n \in N$ , in $N$ ,

$(n + 1) C_{n} = (\binom{2 n}{n}) .$

The identity is that of $(n + 1) C_{n} = (\binom{2 n}{n})$ ; what is new is the route. It is obtained here by extracting a coefficient from the closed form $2 x C = 1 - (1 - 4 x)^{1 / 2}$ ( $2 x C (x) = 1 - (1 - 4 x)^{1 / 2}$ , where $(1 - 4 x)^{1 / 2}$ is the unique square root with constant coefficient $1$ ), with no bijection, no reflection and no group action: only formal algebra in $Q ⟦ x ⟧$ .

Facts & Assumptions

Given: a natural number $n$ , and the Catalan generating function $C$ .

[F1]

$2 x C = 1 - (1 - 4 x)^{1 / 2}$ in $Q ⟦ x ⟧$ ( $2 x C (x) = 1 - (1 - 4 x)^{1 / 2}$ , where $(1 - 4 x)^{1 / 2}$ is the unique square root with constant coefficient $1$ ).

[F2]

$[x^{m}] C = C_{m}$ for every $m$ , and a natural number written where a rational is expected denotes its image under an injective embedding preserving addition and multiplication (The Catalan generating function $C (x) = \sum_{n \geq 0} C_{n} x^{n}$ in $Q ⟦ x ⟧$ ).

[L1]

For every $k \geq 1$ , $k [x^{k}] (1 - 4 x)^{1 / 2} = - 2 (\binom{2 k - 2}{k - 1})$ in $Q$ ( $[x^{k}] (1 - 4 x)^{1 / 2} = - \frac{2}{k} (\binom{2 k - 2}{k - 1})$ for $k \geq 1$ , and $1$ for $k = 0$ ).

[L2]

$[x^{m}] (f + g) = [x^{m}] f + [x^{m}] g$ , $[x^{m}] (r f) = r [x^{m}] f$ , and $[x^{m}] (x^{k} f) = [x^{m - k}] f$ for $k \leq m$ and $0$ for $k > m$ (Coefficient extraction is $R$ -linear, separates formal series, shifts under multiplication by $x^{k}$ , and converts products to finite convolution).

[L3]

$Q$ is a field, so every nonzero rational is invertible (The rationals form a field).

[L4]

$(n + 1) C_{n} = (\binom{2 n}{n})$ in $N$ ( $(n + 1) C_{n} = (\binom{2 n}{n})$ ).

Proof

technique · direct

1.1

Extract the coefficient at the index $n + 1$ from the left-hand side of [F1]: by [L2], $[x^{n + 1}] (2 x C) = 2 [x^{n}] C = 2 C_{n}$ .

F2L2

1.2

Extract it from the right-hand side: by [L2] the constant series $1$ contributes $0$ at a positive index, so $[x^{n + 1}] (1 - (1 - 4 x)^{1 / 2}) = - [x^{n + 1}] (1 - 4 x)^{1 / 2}$ , and multiplying by $n + 1$ and using [L1] with $k = n + 1$ , which is at least $1$ , gives $(n + 1) [x^{n + 1}] (1 - (1 - 4 x)^{1 / 2}) = 2 (\binom{2 n}{n})$ .

L1L2

2.1

By [F1] the two coefficients of steps 1.1 and 1.2 are equal, so multiplying step 1.1 by $n + 1$ gives $2 (n + 1) C_{n} = 2 (\binom{2 n}{n})$ in $Q$ ; cancelling the nonzero rational $2$ by [L3] gives $(n + 1) C_{n} = (\binom{2 n}{n})$ in $Q$ , and the embedding of [F2] being injective, the same identity holds in $N$ . It is the identity of [L4], now proved a third time. At $n = 0$ it reads $C_{0} = 1$ .

F1F2L3L4step 1.1step 1.2∎

Remarks

What makes this a different route and not a rearrangement. The two earlier derivations count a set twice: once directly and once after a reflection or after a group action. This one never counts anything. It turns the recurrence into an algebraic equation, solves that equation inside $Q ⟦ x ⟧$ , and reads a single coefficient off the solution. The only combinatorial input is the recurrence itself.
Where the three derivations meet. All three end at the same identity in $N$ , and the cycle-lemma derivation ends at $(2 n + 1) C_{n} = (\binom{2 n + 1}{n})$ , whose consistency with this one is proved where it is stated. Agreement of the answers is not evidence that the routes are the same; each spends a different hypothesis, and the remark on routes at the end of this page records which.

Depends on

Used by

Why the Catalan count is proved three times, and how the three statements agree Remark

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

A. Postnikov (notes by A. Lin), MIT 18.212 Algebraic Combinatorics, Spring 2019 (standard reference, not scraped)
D. Guichard, An Introduction to Combinatorics and Graph Theory, §3.5 (standard reference, not scraped)

A third derivation of (n+1) Cn=(2nn)(n+1)Cn​=(n2n​), from the closed form of C(x)C(x)